How to use type check in Kotlin?
up vote
0
down vote
favorite
In Kotlin doc, type check use is
but when I write this code
var a="hello"
if (a is String) print(a)
There is a warning
warning: check for instance is always 'true'
if (a is String) print(a)
^
Thank you very much for all answers.
kotlin typechecking
add a comment |
up vote
0
down vote
favorite
In Kotlin doc, type check use is
but when I write this code
var a="hello"
if (a is String) print(a)
There is a warning
warning: check for instance is always 'true'
if (a is String) print(a)
^
Thank you very much for all answers.
kotlin typechecking
In this code,a
will always be a string. That's what the compiler tells you. But you can use it in situations like this: kotlinlang.org/docs/reference/typecasts.html#smart-casts
– Sergio Tulentsev
Nov 10 at 19:02
Well it is always a string.
– EpicPandaForce
Nov 10 at 20:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Kotlin doc, type check use is
but when I write this code
var a="hello"
if (a is String) print(a)
There is a warning
warning: check for instance is always 'true'
if (a is String) print(a)
^
Thank you very much for all answers.
kotlin typechecking
In Kotlin doc, type check use is
but when I write this code
var a="hello"
if (a is String) print(a)
There is a warning
warning: check for instance is always 'true'
if (a is String) print(a)
^
Thank you very much for all answers.
kotlin typechecking
kotlin typechecking
edited Nov 10 at 21:56
D Manokhin
52418
52418
asked Nov 10 at 18:59
Ray Chakrit
72
72
In this code,a
will always be a string. That's what the compiler tells you. But you can use it in situations like this: kotlinlang.org/docs/reference/typecasts.html#smart-casts
– Sergio Tulentsev
Nov 10 at 19:02
Well it is always a string.
– EpicPandaForce
Nov 10 at 20:28
add a comment |
In this code,a
will always be a string. That's what the compiler tells you. But you can use it in situations like this: kotlinlang.org/docs/reference/typecasts.html#smart-casts
– Sergio Tulentsev
Nov 10 at 19:02
Well it is always a string.
– EpicPandaForce
Nov 10 at 20:28
In this code,
a
will always be a string. That's what the compiler tells you. But you can use it in situations like this: kotlinlang.org/docs/reference/typecasts.html#smart-casts– Sergio Tulentsev
Nov 10 at 19:02
In this code,
a
will always be a string. That's what the compiler tells you. But you can use it in situations like this: kotlinlang.org/docs/reference/typecasts.html#smart-casts– Sergio Tulentsev
Nov 10 at 19:02
Well it is always a string.
– EpicPandaForce
Nov 10 at 20:28
Well it is always a string.
– EpicPandaForce
Nov 10 at 20:28
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
In your example, "hello"
is a String
literal. In Kotlin, even when you omit the type for a variable, its type is inferred. The compiler infers the type for var a
from the initializer expression, and so the type of a
is String
. The warning you are getting means that the expression a
that you check is always a String
.
Your variable declaration is equivalent to var a: String = "hello"
, i.e. the variable may only reference a String
, assigning any other type is not allowed.
For example, if you change the variable declaration to var a: Any = "hello"
, there will be no warning since the variable now may hold an instance of any type, not just a String
.
add a comment |
up vote
0
down vote
I just figured out how to use type check by learning from Swift
open class fruit
class banana: fruit()
fun test( a: fruit )
if (a is banana) print("ok")
test(banana())
That's good, but how does this answer your own question about the warning?
– Erwin Bolwidt
Nov 11 at 8:08
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
In your example, "hello"
is a String
literal. In Kotlin, even when you omit the type for a variable, its type is inferred. The compiler infers the type for var a
from the initializer expression, and so the type of a
is String
. The warning you are getting means that the expression a
that you check is always a String
.
Your variable declaration is equivalent to var a: String = "hello"
, i.e. the variable may only reference a String
, assigning any other type is not allowed.
For example, if you change the variable declaration to var a: Any = "hello"
, there will be no warning since the variable now may hold an instance of any type, not just a String
.
add a comment |
up vote
3
down vote
In your example, "hello"
is a String
literal. In Kotlin, even when you omit the type for a variable, its type is inferred. The compiler infers the type for var a
from the initializer expression, and so the type of a
is String
. The warning you are getting means that the expression a
that you check is always a String
.
Your variable declaration is equivalent to var a: String = "hello"
, i.e. the variable may only reference a String
, assigning any other type is not allowed.
For example, if you change the variable declaration to var a: Any = "hello"
, there will be no warning since the variable now may hold an instance of any type, not just a String
.
add a comment |
up vote
3
down vote
up vote
3
down vote
In your example, "hello"
is a String
literal. In Kotlin, even when you omit the type for a variable, its type is inferred. The compiler infers the type for var a
from the initializer expression, and so the type of a
is String
. The warning you are getting means that the expression a
that you check is always a String
.
Your variable declaration is equivalent to var a: String = "hello"
, i.e. the variable may only reference a String
, assigning any other type is not allowed.
For example, if you change the variable declaration to var a: Any = "hello"
, there will be no warning since the variable now may hold an instance of any type, not just a String
.
In your example, "hello"
is a String
literal. In Kotlin, even when you omit the type for a variable, its type is inferred. The compiler infers the type for var a
from the initializer expression, and so the type of a
is String
. The warning you are getting means that the expression a
that you check is always a String
.
Your variable declaration is equivalent to var a: String = "hello"
, i.e. the variable may only reference a String
, assigning any other type is not allowed.
For example, if you change the variable declaration to var a: Any = "hello"
, there will be no warning since the variable now may hold an instance of any type, not just a String
.
edited Nov 10 at 20:26
answered Nov 10 at 20:24
hotkey
57.9k11168186
57.9k11168186
add a comment |
add a comment |
up vote
0
down vote
I just figured out how to use type check by learning from Swift
open class fruit
class banana: fruit()
fun test( a: fruit )
if (a is banana) print("ok")
test(banana())
That's good, but how does this answer your own question about the warning?
– Erwin Bolwidt
Nov 11 at 8:08
add a comment |
up vote
0
down vote
I just figured out how to use type check by learning from Swift
open class fruit
class banana: fruit()
fun test( a: fruit )
if (a is banana) print("ok")
test(banana())
That's good, but how does this answer your own question about the warning?
– Erwin Bolwidt
Nov 11 at 8:08
add a comment |
up vote
0
down vote
up vote
0
down vote
I just figured out how to use type check by learning from Swift
open class fruit
class banana: fruit()
fun test( a: fruit )
if (a is banana) print("ok")
test(banana())
I just figured out how to use type check by learning from Swift
open class fruit
class banana: fruit()
fun test( a: fruit )
if (a is banana) print("ok")
test(banana())
answered Nov 11 at 7:08
Ray Chakrit
72
72
That's good, but how does this answer your own question about the warning?
– Erwin Bolwidt
Nov 11 at 8:08
add a comment |
That's good, but how does this answer your own question about the warning?
– Erwin Bolwidt
Nov 11 at 8:08
That's good, but how does this answer your own question about the warning?
– Erwin Bolwidt
Nov 11 at 8:08
That's good, but how does this answer your own question about the warning?
– Erwin Bolwidt
Nov 11 at 8:08
add a comment |
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In this code,
a
will always be a string. That's what the compiler tells you. But you can use it in situations like this: kotlinlang.org/docs/reference/typecasts.html#smart-casts– Sergio Tulentsev
Nov 10 at 19:02
Well it is always a string.
– EpicPandaForce
Nov 10 at 20:28